Group identities and inverses

A set is a set. A magma is a set with a binary operator. A semigroup is a magma with an associative binary operator. A monoid has a two-sided identity. And a group has two-sided inverses.

I am wondering about one-sided verses two-sided. Under what conditions is an identity element necessarily two-sided? Under what conditions is an inverse necessarily two-sided? And what are the simplest proofs for these?

Theorems I have so far:

• A magma may have multiple distinct left-identities or multiple distinct right-identities, but can never have a distinct left and right identity. [1: $\forall x. lx=x$. 2: $\forall x. xr=x$. 1 implies that $lr=r$ while 2 implies that $lr=l$. So either $l=r$ or at least one of 1 or 2 is false.]

• Associativity plus the existence of a two-sided inverse is enough to imply that any inverse is two-sided. [If $y$ is the left-inverse of $x$ then $xyx = x(yx)=xi=x$. By associativity, $xyx=(xy)x=x$, which implies that $xy=i$. In other words, $y$ is also the right-inverse of $x$.]

I have a feeling that an associative magma cannot have one-sided identities - but I cannot prove this.

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I recall that any three of {all left inverses exist, all right inverses exists, left identity exists, right identity exists} is sufficient to imply the fourth in any semigroup, and I believe there are semigroups in which any two hold but not the other two. – MJD Apr 25 '12 at 14:36
@MarkDominus: In the absence of a specified identity, it's not even clear what "left inverses exist" and "right inverses exist" mean. It's true that a semigroup which has a left identity and left inverses with respect to that left identity is a group (and similarly for right). – Chris Eagle Apr 25 '12 at 14:39
@ChrisEagle Thanks for this clarification. – MJD Apr 25 '12 at 14:40
As for the question at the end of the OP: for the operation $a*b=b$, everything is a left identity but (assuming there are at least two elements) nothing is a right identity. – Chris Eagle Apr 25 '12 at 14:40

The classic result in this area is that if $G$ is a semigroup with a left identity $e$ and such that every element has a left inverse with respect to $e$, then $G$ is a group. There are proofs of this all over the place, in section 1.1 of Hungerford's Algebra for example. Of course the same holds with "left" replaced with "right" throughout.

Proof: For any $g$ in $G$, we have $(gg^{-1})(gg^{-1})=g(g^{-1}g)g^{-1}=geg^{-1}=gg^{-1}$ Multiplying both sides on the left by $(gg^{-1})^{-1}$, we have $egg^{-1}=e$ and hence $gg^{-1}=e$. Thus $g^{-1}$ is in fact a two-sided inverse of $g$ (with respect to the identity $e$). Furthermore, $ge=g(g^{-1}g)=(gg^{-1})g=eg=g$, and hence $e$ is a two-sided identity.

On the counterexamples side, a fertile structure to look at is any set of at least two elements with the operation $a*b=b$. This is a semigroup in which every element is a left identity, while no element is a right identity. Furthermore, if we fix a left identity $e$, then every element has a right inverse (also $e$) with respect to $e$, while only $e$ has a left inverse (in fact everything is left inverse to it). If we relax the condition "$a$ has a left inverse" to mean "there exists $b$ such that $ba$ is a left identity" (rather than picking a specific identity and sticking to it), then everything is left inverse to everything.

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I think in the second paragraph, you mean that every element has a right inverse with respect to $e$, while only $e$ has a left inverse (otherwise this contradicts the result in your first paragraph). – Ted Apr 25 '12 at 15:41
@Ted, yes, thanks. – Chris Eagle Apr 25 '12 at 15:42
+1 for a clearly written answer. Doesn't completely answer my question though; it's still not clear to me whether an operator simply being associative guarantees that any identity element is two-sided or not. Note that I don't have access to any textbooks; I'm just a random guy in my bedroom trying to learn some basic mathematics. If somebody could get me a brief summary of the essential argument, that would be great. – MathematicalOrchid Apr 25 '12 at 20:50
@MathematicalOrchid: My example has one-sided but not two-sided identities. I'll reproduce Hungerford's argument when I have time. – Chris Eagle Apr 25 '12 at 20:58
@MathematicalOrchid: It's clearly associative (and hence a semigroup, as I claimed): $(a*b)*c=b*c=c=a*c=a*(b*c)$. – Chris Eagle Apr 25 '12 at 21:00

Cancellativity is a strong condition which implies that one-sided identities and inverses are two-sided in semigroups. Let $S$ be a semigroup. We say $S$ is cancellative when $$\begin{eqnarray}ac=bc\implies a=b,\\ca=cb\implies a=b.\end{eqnarray}$$

We say $e\in S$ is idempotent when $e^2=e.$ Let $S$ be a cancellative semigroup.

Fact 1. Let $e\in S$ be idempotent. Then $e$ is a two-sided identity element.

Proof. $ex=e^2x$ and so by cancellativity $x=ex.$ Analogously $x=xe.$

Fact 2. Let $e\in S$ be a two-sided identity element. Any left or right inverse with respect to $e$ in $S$ is a two-sided inverse.

Proof. Let $xy=e.$ Then $xyx=ex=x=xe,$ and so by cancellativity $yx=e.$

Fact 3. Let $e\in S$ be such an element that there exists $x\in S$ such that $ex=x$ or $xe=x.$ Then $e$ is a two-sided identity element in $S.$

Proof. Suppose $ex=x.$ For any $y\in S,$ we have $yx=yex,$ whence by cancellativity $y=ye.$ Therefore $e$ is a right identity element. But then $e=e^2$ so $e$ is idempotent. Thus $e$ is a two-sided identity element. It works analogously for the assumption that $xe=x.$

Note that Fact 3. is much stronger than the statement that one-sided identity elements are automatically two-sided.

In fact, when $S$ is finite, we can prove that it's a actually a group.

Proof. Let $S=\{x_1,\ldots,x_n\}$ and $x\in S.$ Let $L=\{xx_1,\ldots,xx_n\}.$ Then it is easy to see that cancellativity implies $$S=L.$$

Therefore there is $i$ such that $x=xx_i.$ From Fact 3. it follows that $x_i$ is a two-sided identity element in $S$. But also, there must be $j$ such that $x_i=xx_j.$ Therefore $x$ has a right inverse. But then it must also be a left-inverse by Fact 2. Therefore $S$ is a group.

$(\{0,1,2,\ldots\},+)$ is an example of a cancellative monoid which isn't a group. The identity element is $0$ and it's two-sided as the facts above require, but there are no units (invertible elements) except $0.$

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Is the cancellative property related to the operator's Cayley table having no duplicates in a given row or column? – MathematicalOrchid Apr 26 '12 at 9:40
Yes, cancellation is equivalent to both rows and columns being injective. There are also one-sided cancellative properties of which one is equivalent to rows being injective and the other to columns being injective. (I'm not saying which is equivalent to which because it depends on how one writes one's Cayley table.) Note that all of the proofs except the first one require associativity. I'd be happy to see counterexamples in non-associative magmas. – user23211 Apr 26 '12 at 10:09
@MathematicalOrchid While the rows and columns must be injective in cancellative semigroups, they need not be surjective. This is the case for the last example in my answer. Only the row corresponding to $0$ is surjective there. However, for finite sets injective=surjective and this is why the proof above works. – user23211 Apr 26 '12 at 11:59

You can find your answer in the following article by A. H. Clifford in the Annals of Mathemtics, volume 34 (1933), pages 865-871:

"A System Arising from a Weakened Set of Group Postulates"

http://www.jstor.org/stable/1968703

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Clifford and Preston, Algebraic theory of semigroups, Vol. 1, covers the following facts:

• If a semigroup has a left identity and a right identity, then they are the same and it is a two-sided identity. (Note that no inverses are needed for this fact.)
• [Dickson, 1905] If a semigroup has a left identity $e$ and weak left inverse for every element with respect to $e$, then it is a group. (That is, $e$ turns out to be a right identity and the weak inverses turn out to be uniquetwo-sided inverses.)
• [Weber, 1986, Huntington, 1901] If a semigroup has weak quotients, i.e., for every $a$ and $b$ there exist $x$ and $y$ such that $ax = b$ and $ya = b$, then it is a group.

I think the proofs for all of them are more or less straightforward, and it is instructive to work them out for oneself.

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Although much of facts have been written down. Just as a summary, proof of all can be found in paper "A System Arising from a Weakened Set of Group Postulates" mentioned by user83548

Let (G, *) be a semigroup with a non empty set E consisting of left identities of G.

1. for all g in G and for all e in E there exist h in h such that h*g=e, then it is group

2. For all g in G and all e in E there exist h in G such that g*h = e, then it not necessary a group.

3. for all g in G there exist e in E and there exist h in G such that h*g=e, then G may not be group(See difference between 2. and 3.and 5.)

4. for all g in G there exist e in E and there exist h in G such that g*h=e, then G may not be group.

5. there exist e(fixed) in E such that for all g in G there exist h in G such that h*g=e then G is group. (problem from I.N. Herstein)

6. there exist e(fixed) in E such that for all g in G there exist h in G such that g*h=e then G need not be group(See difference between 2. and 3.)(problem from Herstein)

7. If E is singleton then with conditions of 3. G is a group.

Note: (a) 2. 3. 4. and 6. are equivalent conditions under given hypothesis(objective of the paper) which make an algebric structure called 'Multiple groups'

Example of multiple group which is not group is easy to construct(See paper for reference if not able to construct)

(b) 5. with given conditions is taken as definition of group in 'Algebra by Van der Waerden'. But make sure to differentiate between 2. and 5. because just noting this difference A. H. CLIFFORD discovered these facts.

(c) About 7. you need to work out a little bit. for reference read index of multiple group in paper given.

(d) And final important fact which completely arrange all facts in one direction is: 2. 3. 4. and 6. are equivalent to: for all a, b in G the equation xa=b has unique solution(under given hypothesis) which implies left cancellation law.

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Sorry don't know how to write 'for every' and 'there exist' and 'belongs to' symbol. If someone can please modify answer with phrases replaced by their symbols, answer will become more presentative and clearer – Sushil Jan 29 '15 at 13:36